How to grep repeated strings on a single line? - grep

Question

I have this a file.txt with one line, whose content is
/app/jdk/java/bin/java -server -Xms3g -Xmx3g -XX:MaxPermSize=256m -Dweblogic.Name=O2pPod8_mapp_msrv1_1 -Djava.security.policy=/app/Oracle/Middleware/Oracle_Home/wlserver/server/lib/weblogic.policy -Djava.security.egd=file:/dev/./urandom -Dweblogic.ProductionModeEnabled=true -Dweblogic.system.BootIdentityFile=/app/Oracle/Middleware/Oracle_Home/user_projects/domains/O2pPod8_domain/servers/O2pPod8_mapp_msrv1_1/data/nodemanager/boot.properties -Dweblogic.nodemanager.ServiceEnabled=true -Dweblogic.nmservice.RotationEnabled=true -Dweblogic.security.SSL.ignoreHostnameVerification=false -Dweblogic.ReverseDNSAllowed=false -Xms8192m -Xmx8192m -XX:MaxPermSize=2048m -XX:NewSize=1300m -XX:MaxNewSize=1300m -XX:SurvivorRatio=4 -XX:TargetSurvivorRatio=90 -XX:+UseParNewGC -XX:+UseConcMarkSweepGC -XX:+CMSParallelRemarkEnabled
and when I do
cat file.txt | grep -io "Xms.*" | awk '{FS" "; print $1} ' | cut -d "s" -f2
output:
3g
why is grep not reading the second occurrence, i.e. I expect 3g and 8192m.
Infact, how do I print only 8192m in this case?

Answer 1

Your regex just says "find Xms followed by anything repeated 0 to n times". That returns the rest of the row from Xms onward.
What you actually want is something like "find Xms followed by anything until there's a whitespace repeated 0 to n times".
grep -io "Xms[^ ]*" file.txt | awk '{FS" "; print $1} ' | cut -d "s" -f2
In [^ ] the ^ means "not"

Answer 2

I'm not really sure what you are trying to achieve here but if you want the endings of all space-separated strings starting with -Xms, using bare awk is:
$ awk -v RS=" " '/^-Xms/{print substr($0,5)}' file
3g
8192m
Explained:
$ awk -v RS=" " ' # space separated records
/^-Xms/ { # strings starting with -Xms
print substr($0,5) # print starting from 5th position
}' file
If you wanted something else (word repeated in the title puzzles me a bit), please update the question with more detailed requirements.
Edit: I just noticed how do I print only 8192m in this case (that's the repeated maybe). Let's add a counter c and not print the first instance:
$ awk -v RS=" " '/^-Xms/&&++c>1{print substr($0,5)}' file
8192m

Answer 3

You could use grep -io "Xms[0-9]*[a-zA-Z]" instead of grep -io "Xms.*" to match a sequence of digits followed by a single character instead the entire line within a single group:
cat file.txt | grep -io "Xms[0-9]*[a-zA-Z]" | awk '{FS" "; print $1} ' | cut -d "s" -f2
Hope this helps!

Answer 4

The .* in your regexp is matching the rest of the line, you need [^ ]* instead. Look:
$ grep -o 'Xms.*' file
Xms3g -Xmx3g -XX:MaxPermSize=256m -Dweblogic.Name=O2pPod8_mapp_msrv1_1 -Djava.security.policy=/app/Oracle/Middleware/Oracle_Home/wlserver/server/lib/weblogic.policy -Djava.security.egd=file:/dev/./urandom -Dweblogic.ProductionModeEnabled=true -Dweblogic.system.BootIdentityFile=/app/Oracle/Middleware/Oracle_Home/user_projects/domains/O2pPod8_domain/servers/O2pPod8_mapp_msrv1_1/data/nodemanager/boot.properties -Dweblogic.nodemanager.ServiceEnabled=true -Dweblogic.nmservice.RotationEnabled=true -Dweblogic.security.SSL.ignoreHostnameVerification=false -Dweblogic.ReverseDNSAllowed=false -Xms8192m -Xmx8192m -XX:MaxPermSize=2048m -XX:NewSize=1300m -XX:MaxNewSize=1300m -XX:SurvivorRatio=4 -XX:TargetSurvivorRatio=90 -XX:+UseParNewGC -XX:+UseConcMarkSweepGC -XX:+CMSParallelRemarkEnabled
$ grep -o 'Xms[^ ]*' file
Xms3g
Xms8192m
$ grep -o 'Xms[^ ]*' file | cut -d's' -f2
3g
8192m
$ grep -o 'Xms[^ ]*' file | cut -d's' -f2 | tail -1
8192m
or more concisely:
$ sed 's/.*Xms\([^ ]*\).*/\1/' file
8192m

Answer 5

The positive lookbehind of PCRE (the form: (?<=RE1)RE2) can resolve the problem easily:
$ grep -oP '(?<=Xms)\S+' file.txt
3g
8192m
Explains:
-o: show only the part of a line matching PATTERN.
-P: PATTERN is a Perl regular expression.
(?<=Xms)\S+: matches all continuous non-whitespace strings which are just following the string Xms.